Question

From a uniform solid cone of base radius $R$ and height $H$ , a symmetric cone of base radius $\frac{R}{2}$ is removed as shown.
Then the center of mass is shifted by $\frac{H}{n}.$ Find the value of $n$ .

Answer 1

Answer: 56

When symmetric cone is removed then center of mass can be calculated by this relation
$y_{cm}=\frac{m_1{y}_1-m_2{y}_2}{m_1-m_2}$ ........(i)
Where $m_1$ and $m_2$ are masses, $y_1$ and $y_2$ are center of masses for original and removed cone respectively. We take origin ${}^{\prime }{O}^{\prime }\left(0,0\right)$ as coordinate of center of mass of original cone.

$△ABC$ and $△DFG$ are similar triangles, so for $△DFG$ , $\frac{EG}{ED}=\frac{EC}{BG}$ , for $△BEC$
$\frac{R}{2\times H^{\prime }}=\frac{R}{H}$
$\Rightarrow H^{\prime }=\frac{H}{2}$
So, center of mass $\left(y_2\right)$ for smaller cone is
$\frac{1}{4}\left(\frac{H}{2}\right)=\frac{H}{8}$
For same density, mass of cone depends on its volume.
Mass of bigger cone is $m_1$ (let) $=\frac{1}{3}\pi R^{2}H$
Mass of smaller cone is $m_2$ (let) $=\frac{1}{3}\pi {\left(\frac{R}{2}\right)}^2\times \frac{H}{2}$
$\Rightarrow m_2=\frac{1}{8}\left(\frac{1}{3}{\left(\pi R\right)}^{2}H\right)=\frac{m_1}{8}$
Putting values of $m_1$ , $y_1=0$ and $m_2=\frac{m_1}{8},y_2=\frac{H}{8}$ in equation (i)
$y_{cm}=\frac{m_1\times 0-\frac{m_1}{8}\times \frac{H}{8}}{m_1-\frac{m_1}{8}}$
Put, $y_{cm}=\frac{H}{n}$ in above equation, in above equation,
$\frac{H}{n}=\frac{H}{56}$
$\Rightarrow n=56$